Solving Shear Force and Bending Moment at Reaction Point R

[fredmahy29]

TL;DR Summary

Bending cantilevered beam when subject to offset transverse force

Hello

Please see attached - trying to figure shear force and bending moment at reaction point R, with a horizontal force applied at top of Z. This is a veranda deck with balustrade.

If F was vertical, then straightforward with SF = F and M = FX. But when F is horizontal as shown, I'm having some difficulty and can't work out the SF or BM diagrams. Perhaps M at R is FZ?

I'd appreciate anything to point me in the right direction to find SF and M at R.
Thanks for your time.
Fred

 

[PhanthomJay]

Summary: Bending cantilevered beam when subject to offset transverse force

Hello

Please see attached - trying to figure shear force and bending moment at reaction point R, with a horizontal force applied at top of Z. This is a veranda deck with balustrade.

If F was vertical, then straightforward with SF = F and M = FX. But when F is horizontal as shown, I'm having some difficulty and can't work out the SF or BM diagrams. Perhaps M at R is FZ?

I'd appreciate anything to point me in the right direction to find SF and M at R.
Thanks for your time.
Fred

Yes the fixed end moment at R is FZ, ccw, or negative by most beam conventions. Now sum forces in y direction to get the vertical reaction at R. What do you get?

 

[fredmahy29]

Yes the fixed end moment at R is FZ, ccw, or negative by most beam conventions. Now sum forces in y direction to get the vertical reaction at R. What do you get?

Thanks PhantomJay
But this is where I'm stuck. To me, reaction R = 0 (surely it couldn't be that R = F?), which is counter-intuitive. Another clue perhaps?

 

[PhanthomJay]

For equilibrium in accordance with Newton’s first law, the sum of all external forces acting on the beam in the y direction , including the applied forces and reaction support forces in the y direction, must equal 0. Same with forces in the x direction. So, you appear to be on track, what is the component of R in the y direction and what is the component of R in the x direction? How much is the shear in the beam at R and along the length of the beam?

 

[fredmahy29]

For equilibrium in accordance with Newton’s first law, the sum of all external forces acting on the beam in the y direction , including the applied forces and reaction support forces in the y direction, must equal 0. Same with forces in the x direction. So, you appear to be on track, what is the component of R in the y direction and what is the component of R in the x direction? How much is the shear in the beam at R and along the length of the beam?

OK thanks. So now the way I see it ...

Horz reaction at Rh = -F
Vert reaction at R can only be due to gravity … something like Rv = -(beam weight*beam length + vertical [baluster] weight) = shear at R

Any closer?
Thanks again for all your help :)

 

[PhanthomJay]

OK thanks. So now the way I see it ...

Horz reaction at Rh = -F
Vert reaction at R can only be due to gravity … something like Rv = -(beam weight*beam length + vertical [baluster] weight) = shear at R

Any closer?
Thanks again for all your help :)

The weight of the horizontal and vertical beams is not given, so you can assume they are negligible and thus equal to 0. (If they were given, gravity acts down on the beam, so the support vertical reaction at R on the beam would have to act upward, or in the positive direction, by convention. ).

So, Rv is 0, Rh is -F (acting on the beam to the left), and the support fixed end moment at R is -Fz (ccw on the beam).

Can you now draw the shear diagram along the length of the beam? (Hint: the Moment and Fx at the support and other end of the horizontal beam do not contribute to the shear in the horizontal beam).

 

[fredmahy29]

Yes, I understand 1st and 2nd para results (although I thought since beam would hog with weight the moment at R would be negative by convention). And I'm afraid I can't find any shear force along the beam!

So obviously I still haven't grasped the concept! Thanks for bearing with me on this!

 

[PhanthomJay]

Yes, I understand 1st and 2nd para results (although I thought since beam would hog with weight the moment at R would be negative by convention). And I'm afraid I can't find any shear force along the beam!
View attachment 249526
So obviously I still haven't grasped the concept! Thanks for bearing with me on this!

Your SFD is correct! There is no shear at all in the horizontal beam, it just carries bending moment and axial tension load.
But your moment diagram is not correct. You started off ok with the negative moment -Fz at the left end of the beam and yes it is a hogging moment. But then the moments along the beam do not slope upward to zero at the right end. Here are a few hints from the calculus of beam theory , dM/dx=V where M is the Moment and V is the Shear:
1. The calculus notes that the slope of the Moment diagram at a point along the beam is the Shear at that point. Since the Shear is 0 at all points, the slope of the Moment diagram is 0 at all points...which implies a flat horizontal line for the Moment diagram, right?
2. Conversely, the calculus notes that the area of the shear diagram between 2 points is the change in Moment between those points. Since the shear diagram has no area, then there is no change in Moment between those points, so the Moment stays constant at -Fz along the full length of the beam.
3. Free Body Diagrams (FBD) are extremely useful. When you take a FBD of the vertical member, note that there is ccw moment -Fz and a force -F at its bottom, for equilibrium. Using Newton's 3rd law, this implies a positive cw moment Fz and force F at the right end of the horizontal beam.

Can you now try to do the Moment diagram again?

 

[fredmahy29]

OK thanks again. So the (horizontal beam) BMD starts off -ve at the cantilever fixing point and continues unchanged to the RHS of the beam; the fixing moment being anticlockwise which balances the clockwise moment at the RHS of beam. This time perhaps?

 

[PhanthomJay]

OK thanks again. So the (horizontal beam) BMD starts off -ve at the cantilever fixing point and continues unchanged to the RHS of the beam; the fixing moment being anticlockwise which balances the clockwise moment at the RHS of beam. This time perhaps?

Yup, that’s good! Always draw the shear diagram first before drawing the Moment diagram. It makes things easier.

 

[fredmahy29]

Well, PhanthomJay, thank you again for all your timely help - far more than I expected! I've learned a little more.
Most appreciated.
Kind regards, Fred

 

[PhanthomJay]

Well, PhanthomJay, thank you again for all your timely help - far more than I expected! I've learned a little more.
Most appreciated.
Kind regards, Fred

You are very welcome, Fred, I am glad I could be of assistance.